<span>PbO
Let's look at each of the 4 compounds and see what's needed.
PbO.
* Oxygen has a valance shell that's missing 2 electrons and wants to get those 2 elections. Lead donates them, so you have a Lead (II) ions. This is a correct choice.
PbCl4
* Chlorine wants to grab 1 electron to fill it's valance shell and Lead donates that election. However, there's 4 chlorine atoms and every one of them wants and electron, and lead is donating all 4 of the desired electrons making the Lead (IV) ion. So this is a bad choice.
Pb2O
* Oxygen still wants 2 electrons and gets them from the lead. But there's 2 lead atoms and each of them donates 1 election making for 2 Lead(I) ions. So this too is a bad choice.
Pb2S
* Sulfur is in the same column of the periodic table as oxygen and if this compound were to exist would have similar properties as Pb2O and would have Lead(I) ions. So this is a bad choice.</span>
Answer:6.719Litres of Cl2 gas.
Explanation:According to eqn of rxn
2Na +Cl2=2NaCl
P=689torr=689/760=0.91atm
T=39°C+273=312K
according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl
But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW
MW of NaCl=23+35.5=58.5g/mol
n=28g(mass given of NaCl)/58.5
n=0.479moles of NaCl
Going back to the reaction,
if 1moles of Cl2 produces 2moles of NaCl
x moles of Cl2 will give 0.479moles of NaCl.
x=0.479*1/2
x=0.239moles of Cl2.
To find the volume, we use ideal ggas eqn,PV=nRT
V=nRT/P
V=0.239*0.082*312/0.91
V=6.719Litres
Answer: 2.71 moles of solute for every 1 kg of solvent.
Explanation: As you know, the molality of a solution tells you the number of moles of solute present for every 1 kg of the solvent.This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.500.mL⋅1.00 g1mL=500. g Next, use the molar mass of the solute to determine how many moles are present in the sample.115g⋅1 mole NanO385.0g=1.353 moles NaNO3So, you know that this solution will contain 1.353moles of sodium nitrate, the solute, for 500. g of water, the solvent.In order to find the molality of the solution, you must figure out how many moles of solute would be present for 1 kg=103g of water.103g water⋅1.353 moles NaNO3500.g water=2.706 moles NaNO3You can thus say that the molality of the solution is equal to molality=2.706 mol kg−1≈2.71 mol kg−1 The answer is rounded to three sig figs.
